Math 814 HW 3. October 16, p. 54: 9, 14, 18, 24, 25, 26

Transcription

1 Math 814 HW 3 October 16, 2007 p. 54: 9, 14, 18, 24, 25, 26 p.54, Exercise 9. If T z = az+b, find necessary and sufficient conditions for T to cz+d preserve the unit circle. T preserves the unit circle iff ae iθ + b = ce iθ + d, for all θ [0, 2π). Squaring both sides and multiplying out, we get a 2 + b 2 + a be iθ + ābe iθ = c 2 + d 2 + c de iθ + cde iθ. Comparing coefficients, we get two equations: a 2 + b 2 = c 2 + d 2, a b = c d. Dividing the first by d 2 and using the second, we get which can be written If the numerators are zero, this means If the numerators are nonzero, we have c 2 b 2 + b 2 d 2 = c 2 d 2 + 1, c 2 b 2 b 2 = c 2 b 2 d 2. c = b, hence a = d. b = d, hence a = c. 1

2 In the first case, there are u, v with u = v = 1 such that c = ub, d = va. Then we have so a b = c d = ub vā, ub b = vā a. This last number, call it λ, also has λ = 1, and we have [ ] [ ] [ ] a b 1 0 a b =. (1) c d 0 λ b ā In the second case, there are u, v with u = v = 1 such that d = ub, c = va. This time, we have a b = c d = vaū b, so vū = 1. But ū = u 1, so v = u, and we have ad bc = aub bua = 0, which is illegal for a Möbius transformation. So the second case does not occur, and all such transformations T are of the form (1). p.54, Exercise 14. Let G be the region between two circles inside one another, tangent at the point a. Map G conformally to the open unit disk. The map T z = (z a) 1 sends a to, hence the circles go to lines l, l. If l and l are not parallel, they must meet at some point z C. Then T 1 z would be a point on both circles other than a. Since there is no such point, the lines l, l are parallel. The original region G meets both circles, so T sends G to the region G 2 between the parallel lines l and l. Next Rz = cz + d be a product of a translation, a rotation and a dilation sending G 2 to the region G 3 = {z C : Im(z) < π/2}. The function e z sends the vertical line segment {z = k + iy, y < π/2} to the semicircle of radius e k in the half-plane G 4 = {z C : Re(z) > 0}. Hence e z maps G 3 conformally onto G 4. Finally, the map Sz = z 1 z + 1 2

3 sends the imaginary axis ir to the circle {z : z = 1} and the point 1 G 4 to the the point 0. Hence S maps G 4 conformally onto the disk {z : z < 1}. p.54, Exercise 18. Refer to the diagram on page 55. Since M(ia) = 0, the regions B, C, E, F which touch ia are sent, in some order, to the regions U, V, X, Y which touch 0. Since M(ib) =, the regions B, E, which touch ib, are sent, in some order, to the regions U, X, which touch. It follows that C, F go to V, Y, in some order. To determine which region goes to which, let x, y be small positive real numbers, so that the point z = x + iy + ia belongs to E. Then the imaginary part of Mz is a positive number times x(b a), hence is positive. It follows that ME = U, and hence MB = X. Since B, C meet in a line, so does X, MC. It follows that MC = Y, so MF = V. Since A, B meet in a line, so does MA, X. Hence MA = Z and finally MD = W. To summarize: A Z B X C Y D W E U F V. p.54, Exercise 24. Let T be a Möbius transformation with T I. Show that if S is another Möbius transformation with the same fixed points as T, then S commutes with T. By conjugating in the Möbius group, we may assume T =. This means T z = az + b for some a, b C with a 0. Since S also fixes, we have Sz = cz + d for some c, d C with c 0. Suppose a = 1. Then T has as its only fixed-point. Conversely, since S also has as its only fixed point, it follows that c = 1. Hence S, T are translations, which commute with each other. (In fact ST z = z + b + d = T Sz.) Suppose a 1. Then T has a second fixed-point z 0 = b/(1 a). The translation Rz = z z 0 fixes and sends z 0 0. Conjugating T and S by R, we may assume z 0 = 0. Hence T, S belong to the subgroup fixing 0 and. Hence T z = az and Sz = cz, so S and T commute. 3

4 p.54, Exercise 25. Find all abelian subgroups of the group M of Möbius transformations. Let A be an abelian subgroup of M and let I T A. As in the previous problem, conjugate so that T =. Suppose that is the only fixed-point of T. For any S A, we have T S = ST = S, so S is also fixed by T. By uniqueness of the fixed-point, we have S =. If w is any fixed-point of S, then ST w = T Sw = T w. Since S has at most two fixed-points, this implies that T w = w or T w =. Either way, we have w =. Hence is the unique fixed-point of any nonidentity element in A. It follows that A is contained in the subgroup of translations, which is isomorphic to C. Suppose now that T has another fixed-point besides. As in the previous problem, we may assume this other fixed-point is 0, so that T z = az. If a 1, You can check directly that the only Möbius transformations commuting with T are of the form cz. Hence A is contained in the group of dilations and rotations, which is isomorphic to C (the multiplicative group of nonzero complex numbers). Alternatively, the previous arguments show that if I S A then S{0, } = {0, }. If S switches 0 and, then Sz = c/z. But then we would have ST S 1 = T 1 T, since a 1. So S must fix both 0,, hence (again) is of the form Sz = cz. Finally, suppose T z = z. Then any Sz = c/z commutes with T. Conjugating by cz, we can assume c = 1. Hence T is contained in the abelian subgroup {z, z, 1, 1 }, which is Klein s Viergruppe. z z p.54, Exercise 26. Let ϕ : GL 2 (C) M be the map given by ( ) a b ϕ = az + b c d cz + d. You can check directly that if ( ) ( ) a b a A =, and A b = c d c d, 4

5 then ϕ(aa ) = ϕ(a) ϕ(a ). So ϕ is a homomorphism. If az + b cz + d = z for all z, then az + b = z(cz + d). Comparing coefficients shows that c = b = 0 and a = d. So the kernel of ϕ is the subgroup Z = {ai : a C } of GL 2 (C). More conceptually, the homomorphism ϕ arises from the action of GL 2 (C) on lines in C 2. If a matrix preserves every line, then every line is an eigenvector, which means that A is a scalar matrix. Note incidentally that the fixed points of ϕ(a), which we studied in the previous problems, are exactly the eigenlines of A in CP 1. The restriction of ϕ to SL 2 (C) is still surjective, because for any A GL 2 (C), both A and (det A) 1/2 A have the same image under ϕ and the latter matrix has determinant = 1. The kernel of this restriction is then Z SL 2 (C) = {±I}. So SL 2 (C) is a two-fold cover of M. There is a trap here, which, if you ever study algebraic groups, you will tumble into. Namely, not every real Möbius transformation (i.e., with a, b, c, d R) comes from a matrix in SL 2 (R). I ll let you puzzle that out (hint: z). 5